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Maths Question 16 – JEE-MAIN 2025

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If y(x)=|sin⁡xcos⁡xsin⁡x+cos⁡x+1272827111|, x∈R, then d2ydx2+y is equal to

Look for column or row operations that can simplify the determinant before expansion, especially if elements are sums of other elements.

Step 1: Simplify the Determinant for y(x)✦ Active

First, simplify the given determinant for y(x) by applying column operations. Apply the operation C3→C3−C1−C2 to the determinant:

y(x)=|sin⁡xcos⁡xsin⁡x+cos⁡x+1272827111|=|sin⁡xcos⁡x1272827−27−28111−1−1|=|sin⁡xcos⁡x12728−2811−1|

Now, expand this simplified determinant (e.g., along the third row R3):

y(x)=1⋅((cos⁡x)(−28)−(1)(28))−1⋅((sin⁡x)(−28)−(1)(27))+(−1)⋅((sin⁡x)(28)−(cos⁡x)(27)) =(−28cos⁡x−28)−(−28sin⁡x−27)−(28sin⁡x−27cos⁡x) =−28cos⁡x−28+28sin⁡x+27−28sin⁡x+27cos⁡x =(−28cos⁡x+27cos⁡x)+(28sin⁡x−28sin⁡x)+(−28+27) =−cos⁡x−1
Step 2: Calculate the First and Second Derivatives○ Expand

Now that we have y(x)=−1−cos⁡x, we can find its first and second derivatives:

dydx=ddx(−1−cos⁡x)=0−(−sin⁡x)=sin⁡x d2ydx2=ddx(sin⁡x)=cos⁡x
Step 3: Compute the Final Expression○ Expand

Finally, substitute the expressions for y(x) and d2ydx2 into the required expression d2ydx2+y:

d2ydx2+y=cos⁡x+(−1−cos⁡x) =cos⁡x−1−cos⁡x =−1
💡 Teacher's Secret Hint

Ensure all terms cancel out correctly after substitution.

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